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exis [7]
3 years ago
10

How does math relate to electrons and ion formation?

Chemistry
2 answers:
Elenna [48]3 years ago
7 0
In order to find the NET nuclear charge from an atom's valence electron to the proton nucleus, you need to do some simple math in order to find the charge.

You would also need to apply math when you need to figure out what the charge on an ion is by either adding or taking away electrons depending on whether it is an anion or cation.
AlladinOne [14]3 years ago
5 0

Answer and explanation;

Ions are formed when an atom looses or gains electrons.

-Ionic charges are created when there is an imbalance of protons and electrons on an atom. Since we can not change the number of protons on the nucleus of an atom, a positively charged ion will be created when there are fewer electrons than protons (in other words when an atom looses electron). A negatively charged ion is formed or created when there are more electrons than protons in an atom (when an atom gains electrons).

-When electron are added, you are adding more negative charge and thus, results in the creation of a negative ion (anion), and when electrons are being lost by an atom, one is removing negative charge, resulting in creation of a positive ion (cation).

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

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